[leetcode] Valid Palindrome @ Python

原题地址：https://oj.leetcode.com/problems/valid-palindrome/

题意：

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

For example,
"A man, a plan, a canal: Panama" is a palindrome

"radar" is a palindrome

"rotator" is a palindrome
"race a car" is not a palindrome.

Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.

解题思路1：利用python的list comprehension 以及string 的　isalnum()函数我们可以写出极为简短的解决方法　（原创）：

class Solution:
    # @param s, a string
    # @return a boolean
    def isPalindrome(self, s):
        newS= [i.lower() for i in s if i.isalnum()]
        #return newS == newS[::-1]
        return newS[:len(newS)/2] == newS[(len(newS)+1)/2:][::-1]　

需要说明的是程序的最后一行对于newS长度为奇或者为偶都通用

newS[:len(newS)/2] == newS[(len(newS)+1)/2:][::-1]

解题思路2：将不是字母的字符去掉，然后转换成小写，然后简单的回文判断。

class Solution:
    # @param s, a string
    # @return a boolean
    def isPalindrome(self, s):
        if s == '':
            return True
        else:
            sTmp = ''
            for i in range(0, len(s)):
                if s[i] >= 'a' and s[i] <= 'z' or s[i] >= '0' and s[i] <= '9' or s[i] >= 'A' and s[i] <= 'Z':
                    sTmp += s[i]
            sTmp = sTmp.lower()
            for i in range(0, len(sTmp)/2):
                if sTmp[i] != sTmp[len(sTmp)-1-i]:
                    return False
            return True

 参考致谢：

［１］http://www.cnblogs.com/zuoyuan/p/3765882.html